Answers:
a. 1.80 s
b. 0.363 m/s (10.4 cm)
0.590 m/s (16.9 cm)
c. 1.27 m/s² (10.4 cm)
2.06 m/s² (16.9 cm)
d. 0.00997 N (10.4 cm)
0.0162 N (16.9 cm)
Explanation:
Part a.
First, let's convert 33 1/3 rpm to rad/s, so
[tex]33.333\text{ rev/min}\times\frac{2\pi\text{ rad}}{1\text{ rev}}\times\frac{1\text{ min}}{60\text{ s}}=3.49\text{ rad/s}[/tex]
So, the angular velocity w of the record is 3.49 rad/s. Now, we can calculate the period T as follows
[tex]T=\frac{2\pi}{w}=\frac{2\pi}{3.49\text{ rad/s}}=1.80\text{ s}[/tex]
Part b.
The speed of the coin can be calculated as
v = wr
Where r is the radius, so the speed at r = 10.4 cm and at r = 16.9 cm are equal to
[tex]\begin{gathered} v=(3.49\text{ rad/s)(0.104 m) = }0.363\text{ m/s } \\ v=(3.49\text{ rad/s)(0.169 m) = 0.590 m/s} \end{gathered}[/tex]
Part c.
Then, the acceleration is equal to
[tex]a=\frac{v^2}{r}[/tex]
So, for each case, we get
[tex]\begin{gathered} a=\frac{(0.363m/s)^2}{0.104\text{ m}}=1.27m/s^2 \\ a=\frac{(0.590m/s)^2}{0.169\text{ m}}=2.06m/s^2 \end{gathered}[/tex]
Part d.
Finally, the net force is equal to the mass times the acceleration, so for each case, we get
[tex]\begin{gathered} F_{\text{net}}=ma \\ F_{\text{net}}=(0.00787\text{ kg)}(1.27m/s^2)=0.00997\text{ N} \\ F_{\text{net}}=(0.00787\text{ kg)}(2.06m/s^2)=0.0162\text{ N} \end{gathered}[/tex]
Therefore, the answers are:
a. 1.80 s
b. 0.363 m/s (10.4 cm)
0.590 m/s (16.9 cm)
c. 1.27 m/s² (10.4 cm)
2.06 m/s² (16.9 cm)
d. 0.00997 N (10.4 cm)
0.0162 N (16.9 cm)
